Problem: Multiply the following complex numbers: $({3}) \cdot ({2-i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3}) \cdot ({2-i}) = $ $ ({3} \cdot {2}) + ({3} \cdot {-1}i) + ({0}i \cdot {2}) + ({0}i \cdot {-1}i) $ Then simplify the terms: $ (6) + (-3i) + (0i) + (0 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 6 + (-3 + 0)i + 0i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 6 + (-3 + 0)i - 0 $ The result is simplified: $ (6 - 0) + (-3i) = 6-3i $